Question

Which of the following elements is unable to form \[{\text{MF}}_6^{3 - }\] ions?
A.B
B.Ga
C.In
D.Al

Answer 1

Hint: The smaller the size of atom or element is, the more difficult it would be to form bonds with 6 atoms due to steric hindrance. The one which is smallest in size among the all will not be able to form the complex. The non availability of d orbital makes the element unfit for formation of octahedral complexes.

Complete step by step answer:
All the above elements in the given option belong to the same group that is group number 13 of the periodic table and is known as boron family. Since fluorine has an oxidation state as $ - 1$ , so the oxidation state of the metal atom should be $ + 3$ in order to form the above complex.
B is a chemical symbol for boron. The atomic number of boron is 5 and it belongs to the second period. In the second period only s and p orbital are available. S orbital can have a maximum of two electrons and p orbital can have a maximum of 6 electrons. The electronic configuration of boron is:
\[1{{\text{s}}^2}2{{\text{s}}^2}2{{\text{p}}^1}\].
The electronic configuration of \[{{\text{B}}^{ + 3}}\] is \[1{{\text{s}}^2}2{{\text{s}}^0}2{{\text{p}}^0}\]. So the boron can only take up 4 pairs of electrons and can form a complex \[{\text{BF}}_4^ - \] and not \[{\text{BF}}_6^{3 - }\].
All the other elements of this group such as aluminium, Gallium and Indium belong to period number 3, 4 and 5 respectively and have d orbital to accept more electrons unlike boron. So boron is the only element that does not have the capability to form the complex.

Hence the correct option is A.

Note:
As we can see that the given complex has 6 fluorine ions attached to the central metal. Such complexes are called octahedral complexes. Also the availability of d orbital only determines the capability of an element to make the required complex and does not confirm the formation of the complex, because other factors are also needed to be considered.